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Chapter 1

Q. 1.3

Find the gravitational force exerted by the earth on a 70-kg man whose elevation above the surface of the earth equals the radius of the earth. The mass and radius of the earth are M_{e} = 5.9742 × 10^{24}\:kg and R_{e} = 6378\:km, respectively.

Step-by-Step

Verified Solution

Consider a body of mass m located at the distance 2R_{e} from the center of the earth (of mass M_{e}). The law of universal gravitation, from Eq. (11.4), states that the body is attracted to the earth by the force F given by

F=G\frac{mM_{e}}{(2R_{e})^{2}}

where G = 6.67 × 10^{−11}\:m^{3}/(kg·s^{2}) is the universal gravitational constant. Substituting the values for G and the given parameters, the earth’s gravitational force acting on the 70-kg man is

F=(6.67\times 10^{-11})\frac{(70)(5.9742\times10^{24})}{[2(6378\times 10^{3})]^{2}}=171.4\:N